Find a specific file in the nearest ancestor of the current working directory

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP












3















I'd like to find a way to find a given file by looking upward in the directory structure, rather than recursively searching through child directories.



There's a node module that appears to do exactly what I want, but I don't want to depend on installing JavaScript or packages like that. Is there a shell command for this? A way to make find do that? Or a standard approach that I just wasn't able to find through Googling?










share|improve this question
























  • Do you also want this behavior of that node module? "If X/.dir/file.ext exists, return it." (that searches an immediate sub directory of X, not purely ancestors)

    – Jeff Schaller
    Jul 2 '16 at 14:27















3















I'd like to find a way to find a given file by looking upward in the directory structure, rather than recursively searching through child directories.



There's a node module that appears to do exactly what I want, but I don't want to depend on installing JavaScript or packages like that. Is there a shell command for this? A way to make find do that? Or a standard approach that I just wasn't able to find through Googling?










share|improve this question
























  • Do you also want this behavior of that node module? "If X/.dir/file.ext exists, return it." (that searches an immediate sub directory of X, not purely ancestors)

    – Jeff Schaller
    Jul 2 '16 at 14:27













3












3








3


0






I'd like to find a way to find a given file by looking upward in the directory structure, rather than recursively searching through child directories.



There's a node module that appears to do exactly what I want, but I don't want to depend on installing JavaScript or packages like that. Is there a shell command for this? A way to make find do that? Or a standard approach that I just wasn't able to find through Googling?










share|improve this question
















I'd like to find a way to find a given file by looking upward in the directory structure, rather than recursively searching through child directories.



There's a node module that appears to do exactly what I want, but I don't want to depend on installing JavaScript or packages like that. Is there a shell command for this? A way to make find do that? Or a standard approach that I just wasn't able to find through Googling?







shell-script files find directory






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Jul 2 '16 at 13:04









Jeff Schaller

41.2k1056131




41.2k1056131










asked Jul 1 '16 at 23:07









iconoclasticonoclast

3,80163870




3,80163870












  • Do you also want this behavior of that node module? "If X/.dir/file.ext exists, return it." (that searches an immediate sub directory of X, not purely ancestors)

    – Jeff Schaller
    Jul 2 '16 at 14:27

















  • Do you also want this behavior of that node module? "If X/.dir/file.ext exists, return it." (that searches an immediate sub directory of X, not purely ancestors)

    – Jeff Schaller
    Jul 2 '16 at 14:27
















Do you also want this behavior of that node module? "If X/.dir/file.ext exists, return it." (that searches an immediate sub directory of X, not purely ancestors)

– Jeff Schaller
Jul 2 '16 at 14:27





Do you also want this behavior of that node module? "If X/.dir/file.ext exists, return it." (that searches an immediate sub directory of X, not purely ancestors)

– Jeff Schaller
Jul 2 '16 at 14:27










3 Answers
3






active

oldest

votes


















5














This is a direct translation of the find-config algorithm in generic shell commands (tested under bash, ksh, and zsh), where I use a return code of 0 to mean success and 1 to mean NULL/failure.



findconfig() 
# from: https://www.npmjs.com/package/find-config#algorithm
# 1. If X/file.ext exists and is a regular file, return it. STOP
# 2. If X has a parent directory, change X to parent. GO TO 1
# 3. Return NULL.

if [ -f "$1" ]; then
printf '%sn' "$PWD%//$1"
elif [ "$PWD" = / ]; then
false
else
# a subshell so that we don't affect the caller's $PWD
(cd .. && findconfig "$1")
fi



Sample run, with the setup stolen copied and extended from Stephen Harris's answer:



$ mkdir -p ~/tmp/iconoclast
$ cd ~/tmp/iconoclast
$ mkdir -p A/B/C/D/E/F A/good/show
$ touch A/good/show/this A/B/C/D/E/F/srchup A/B/C/thefile
$ cd A/B/C/D/E/F
$ findconfig thefile
/home/jeff/tmp/iconoclast/A/B/C/thefile
$ echo "$?"
0
$ findconfig foobar
$ echo "$?"
1





share|improve this answer

























  • Why is .dir hard-coded into the function?

    – iconoclast
    Jan 22 at 17:20











  • @iconoclast on account of the spec referenced in the question: npmjs.com/package/find-config

    – Jeff Schaller
    Jan 22 at 17:23











  • I don't think the .dir given here is meant to be taken literally. I certainly hope not, because that would make this NPM package not very useful.

    – iconoclast
    Jan 22 at 19:00











  • It's possible I misunderstood, then! I'm not familiar with node.js. Is it your understanding that the code is checking for the file in the/any immediate subdirectory?

    – Jeff Schaller
    Jan 22 at 19:05






  • 1





    it's just checking in the current directory and its parents. So I think we can just remove the second line of code inside the function, and it looks like it should work nicely.—I just updated the code (and tested it first, briefly) but I didn't update the sample run

    – iconoclast
    Jan 22 at 20:15



















3














A simple loop of checking the current directory and if it's not found then strip off the last component would work



#!/bin/bash

wantfile="$1"

dir=$(realpath .)

found=""

while [ -z "$found" -a -n "$dir" ]
do
if [ -e "$dir/$wantfile" ]
then
found="$dir/$wantfile"
fi
dir=$dir%/*
done

if [ -z "$found" ]
then
echo Can not find: $wantfile
else
echo Found: $found
fi


For example, if this is the directory tree:



$ find /tmp/A
/tmp/A
/tmp/A/good
/tmp/A/good/show
/tmp/A/good/show/this
/tmp/A/B
/tmp/A/B/C
/tmp/A/B/C/thefile
/tmp/A/B/C/D
/tmp/A/B/C/D/E
/tmp/A/B/C/D/E/F
/tmp/A/B/C/D/E/F/srchup

$ pwd
/tmp/A/B/C/D/E/F

$ ./srchup thefile
Found: /tmp/A/B/C/thefile


We can see that the search went up the tree until it found what we were looking for.






share|improve this answer






























    3














    One way to do it:



    #! /bin/sh
    dir=$(pwd -P)
    while [ -n "$dir" -a ! -f "$dir/$1" ]; do
    dir=$dir%/*
    done
    if [ -f "$dir/$1" ]; then printf '%sn' "$dir/$1"; fi


    Replace pwd -P by pwd -L if you want to follow symlinks instead of checking physical directories.






    share|improve this answer























    • Any reason you use -a? [ -n "$dir" ] && ! [ -f "$dir/$1" ] is more reliable. (Exercise: Try determining correct parsing for [ -n "$a" -a "$b" = "$c" ] if a='=' and b='-o'.)

      – Wildcard
      Jul 2 '16 at 5:19











    • @Wildcard Because I'm an old fart and set in my bad old ways. :) Also, because it avoids running a second [ on shells where [ is not a builtin. As for [ -n "$a" -a "$b" = "$c" ], that's why you always write x"$b" = x"$c" rather than just "$b" = "$c".

      – Satō Katsura
      Jul 2 '16 at 5:30











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    This is a direct translation of the find-config algorithm in generic shell commands (tested under bash, ksh, and zsh), where I use a return code of 0 to mean success and 1 to mean NULL/failure.



    findconfig() 
    # from: https://www.npmjs.com/package/find-config#algorithm
    # 1. If X/file.ext exists and is a regular file, return it. STOP
    # 2. If X has a parent directory, change X to parent. GO TO 1
    # 3. Return NULL.

    if [ -f "$1" ]; then
    printf '%sn' "$PWD%//$1"
    elif [ "$PWD" = / ]; then
    false
    else
    # a subshell so that we don't affect the caller's $PWD
    (cd .. && findconfig "$1")
    fi



    Sample run, with the setup stolen copied and extended from Stephen Harris's answer:



    $ mkdir -p ~/tmp/iconoclast
    $ cd ~/tmp/iconoclast
    $ mkdir -p A/B/C/D/E/F A/good/show
    $ touch A/good/show/this A/B/C/D/E/F/srchup A/B/C/thefile
    $ cd A/B/C/D/E/F
    $ findconfig thefile
    /home/jeff/tmp/iconoclast/A/B/C/thefile
    $ echo "$?"
    0
    $ findconfig foobar
    $ echo "$?"
    1





    share|improve this answer

























    • Why is .dir hard-coded into the function?

      – iconoclast
      Jan 22 at 17:20











    • @iconoclast on account of the spec referenced in the question: npmjs.com/package/find-config

      – Jeff Schaller
      Jan 22 at 17:23











    • I don't think the .dir given here is meant to be taken literally. I certainly hope not, because that would make this NPM package not very useful.

      – iconoclast
      Jan 22 at 19:00











    • It's possible I misunderstood, then! I'm not familiar with node.js. Is it your understanding that the code is checking for the file in the/any immediate subdirectory?

      – Jeff Schaller
      Jan 22 at 19:05






    • 1





      it's just checking in the current directory and its parents. So I think we can just remove the second line of code inside the function, and it looks like it should work nicely.—I just updated the code (and tested it first, briefly) but I didn't update the sample run

      – iconoclast
      Jan 22 at 20:15
















    5














    This is a direct translation of the find-config algorithm in generic shell commands (tested under bash, ksh, and zsh), where I use a return code of 0 to mean success and 1 to mean NULL/failure.



    findconfig() 
    # from: https://www.npmjs.com/package/find-config#algorithm
    # 1. If X/file.ext exists and is a regular file, return it. STOP
    # 2. If X has a parent directory, change X to parent. GO TO 1
    # 3. Return NULL.

    if [ -f "$1" ]; then
    printf '%sn' "$PWD%//$1"
    elif [ "$PWD" = / ]; then
    false
    else
    # a subshell so that we don't affect the caller's $PWD
    (cd .. && findconfig "$1")
    fi



    Sample run, with the setup stolen copied and extended from Stephen Harris's answer:



    $ mkdir -p ~/tmp/iconoclast
    $ cd ~/tmp/iconoclast
    $ mkdir -p A/B/C/D/E/F A/good/show
    $ touch A/good/show/this A/B/C/D/E/F/srchup A/B/C/thefile
    $ cd A/B/C/D/E/F
    $ findconfig thefile
    /home/jeff/tmp/iconoclast/A/B/C/thefile
    $ echo "$?"
    0
    $ findconfig foobar
    $ echo "$?"
    1





    share|improve this answer

























    • Why is .dir hard-coded into the function?

      – iconoclast
      Jan 22 at 17:20











    • @iconoclast on account of the spec referenced in the question: npmjs.com/package/find-config

      – Jeff Schaller
      Jan 22 at 17:23











    • I don't think the .dir given here is meant to be taken literally. I certainly hope not, because that would make this NPM package not very useful.

      – iconoclast
      Jan 22 at 19:00











    • It's possible I misunderstood, then! I'm not familiar with node.js. Is it your understanding that the code is checking for the file in the/any immediate subdirectory?

      – Jeff Schaller
      Jan 22 at 19:05






    • 1





      it's just checking in the current directory and its parents. So I think we can just remove the second line of code inside the function, and it looks like it should work nicely.—I just updated the code (and tested it first, briefly) but I didn't update the sample run

      – iconoclast
      Jan 22 at 20:15














    5












    5








    5







    This is a direct translation of the find-config algorithm in generic shell commands (tested under bash, ksh, and zsh), where I use a return code of 0 to mean success and 1 to mean NULL/failure.



    findconfig() 
    # from: https://www.npmjs.com/package/find-config#algorithm
    # 1. If X/file.ext exists and is a regular file, return it. STOP
    # 2. If X has a parent directory, change X to parent. GO TO 1
    # 3. Return NULL.

    if [ -f "$1" ]; then
    printf '%sn' "$PWD%//$1"
    elif [ "$PWD" = / ]; then
    false
    else
    # a subshell so that we don't affect the caller's $PWD
    (cd .. && findconfig "$1")
    fi



    Sample run, with the setup stolen copied and extended from Stephen Harris's answer:



    $ mkdir -p ~/tmp/iconoclast
    $ cd ~/tmp/iconoclast
    $ mkdir -p A/B/C/D/E/F A/good/show
    $ touch A/good/show/this A/B/C/D/E/F/srchup A/B/C/thefile
    $ cd A/B/C/D/E/F
    $ findconfig thefile
    /home/jeff/tmp/iconoclast/A/B/C/thefile
    $ echo "$?"
    0
    $ findconfig foobar
    $ echo "$?"
    1





    share|improve this answer















    This is a direct translation of the find-config algorithm in generic shell commands (tested under bash, ksh, and zsh), where I use a return code of 0 to mean success and 1 to mean NULL/failure.



    findconfig() 
    # from: https://www.npmjs.com/package/find-config#algorithm
    # 1. If X/file.ext exists and is a regular file, return it. STOP
    # 2. If X has a parent directory, change X to parent. GO TO 1
    # 3. Return NULL.

    if [ -f "$1" ]; then
    printf '%sn' "$PWD%//$1"
    elif [ "$PWD" = / ]; then
    false
    else
    # a subshell so that we don't affect the caller's $PWD
    (cd .. && findconfig "$1")
    fi



    Sample run, with the setup stolen copied and extended from Stephen Harris's answer:



    $ mkdir -p ~/tmp/iconoclast
    $ cd ~/tmp/iconoclast
    $ mkdir -p A/B/C/D/E/F A/good/show
    $ touch A/good/show/this A/B/C/D/E/F/srchup A/B/C/thefile
    $ cd A/B/C/D/E/F
    $ findconfig thefile
    /home/jeff/tmp/iconoclast/A/B/C/thefile
    $ echo "$?"
    0
    $ findconfig foobar
    $ echo "$?"
    1






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Jan 22 at 22:40









    Stéphane Chazelas

    305k57574928




    305k57574928










    answered Jul 2 '16 at 14:57









    Jeff SchallerJeff Schaller

    41.2k1056131




    41.2k1056131












    • Why is .dir hard-coded into the function?

      – iconoclast
      Jan 22 at 17:20











    • @iconoclast on account of the spec referenced in the question: npmjs.com/package/find-config

      – Jeff Schaller
      Jan 22 at 17:23











    • I don't think the .dir given here is meant to be taken literally. I certainly hope not, because that would make this NPM package not very useful.

      – iconoclast
      Jan 22 at 19:00











    • It's possible I misunderstood, then! I'm not familiar with node.js. Is it your understanding that the code is checking for the file in the/any immediate subdirectory?

      – Jeff Schaller
      Jan 22 at 19:05






    • 1





      it's just checking in the current directory and its parents. So I think we can just remove the second line of code inside the function, and it looks like it should work nicely.—I just updated the code (and tested it first, briefly) but I didn't update the sample run

      – iconoclast
      Jan 22 at 20:15


















    • Why is .dir hard-coded into the function?

      – iconoclast
      Jan 22 at 17:20











    • @iconoclast on account of the spec referenced in the question: npmjs.com/package/find-config

      – Jeff Schaller
      Jan 22 at 17:23











    • I don't think the .dir given here is meant to be taken literally. I certainly hope not, because that would make this NPM package not very useful.

      – iconoclast
      Jan 22 at 19:00











    • It's possible I misunderstood, then! I'm not familiar with node.js. Is it your understanding that the code is checking for the file in the/any immediate subdirectory?

      – Jeff Schaller
      Jan 22 at 19:05






    • 1





      it's just checking in the current directory and its parents. So I think we can just remove the second line of code inside the function, and it looks like it should work nicely.—I just updated the code (and tested it first, briefly) but I didn't update the sample run

      – iconoclast
      Jan 22 at 20:15

















    Why is .dir hard-coded into the function?

    – iconoclast
    Jan 22 at 17:20





    Why is .dir hard-coded into the function?

    – iconoclast
    Jan 22 at 17:20













    @iconoclast on account of the spec referenced in the question: npmjs.com/package/find-config

    – Jeff Schaller
    Jan 22 at 17:23





    @iconoclast on account of the spec referenced in the question: npmjs.com/package/find-config

    – Jeff Schaller
    Jan 22 at 17:23













    I don't think the .dir given here is meant to be taken literally. I certainly hope not, because that would make this NPM package not very useful.

    – iconoclast
    Jan 22 at 19:00





    I don't think the .dir given here is meant to be taken literally. I certainly hope not, because that would make this NPM package not very useful.

    – iconoclast
    Jan 22 at 19:00













    It's possible I misunderstood, then! I'm not familiar with node.js. Is it your understanding that the code is checking for the file in the/any immediate subdirectory?

    – Jeff Schaller
    Jan 22 at 19:05





    It's possible I misunderstood, then! I'm not familiar with node.js. Is it your understanding that the code is checking for the file in the/any immediate subdirectory?

    – Jeff Schaller
    Jan 22 at 19:05




    1




    1





    it's just checking in the current directory and its parents. So I think we can just remove the second line of code inside the function, and it looks like it should work nicely.—I just updated the code (and tested it first, briefly) but I didn't update the sample run

    – iconoclast
    Jan 22 at 20:15






    it's just checking in the current directory and its parents. So I think we can just remove the second line of code inside the function, and it looks like it should work nicely.—I just updated the code (and tested it first, briefly) but I didn't update the sample run

    – iconoclast
    Jan 22 at 20:15














    3














    A simple loop of checking the current directory and if it's not found then strip off the last component would work



    #!/bin/bash

    wantfile="$1"

    dir=$(realpath .)

    found=""

    while [ -z "$found" -a -n "$dir" ]
    do
    if [ -e "$dir/$wantfile" ]
    then
    found="$dir/$wantfile"
    fi
    dir=$dir%/*
    done

    if [ -z "$found" ]
    then
    echo Can not find: $wantfile
    else
    echo Found: $found
    fi


    For example, if this is the directory tree:



    $ find /tmp/A
    /tmp/A
    /tmp/A/good
    /tmp/A/good/show
    /tmp/A/good/show/this
    /tmp/A/B
    /tmp/A/B/C
    /tmp/A/B/C/thefile
    /tmp/A/B/C/D
    /tmp/A/B/C/D/E
    /tmp/A/B/C/D/E/F
    /tmp/A/B/C/D/E/F/srchup

    $ pwd
    /tmp/A/B/C/D/E/F

    $ ./srchup thefile
    Found: /tmp/A/B/C/thefile


    We can see that the search went up the tree until it found what we were looking for.






    share|improve this answer



























      3














      A simple loop of checking the current directory and if it's not found then strip off the last component would work



      #!/bin/bash

      wantfile="$1"

      dir=$(realpath .)

      found=""

      while [ -z "$found" -a -n "$dir" ]
      do
      if [ -e "$dir/$wantfile" ]
      then
      found="$dir/$wantfile"
      fi
      dir=$dir%/*
      done

      if [ -z "$found" ]
      then
      echo Can not find: $wantfile
      else
      echo Found: $found
      fi


      For example, if this is the directory tree:



      $ find /tmp/A
      /tmp/A
      /tmp/A/good
      /tmp/A/good/show
      /tmp/A/good/show/this
      /tmp/A/B
      /tmp/A/B/C
      /tmp/A/B/C/thefile
      /tmp/A/B/C/D
      /tmp/A/B/C/D/E
      /tmp/A/B/C/D/E/F
      /tmp/A/B/C/D/E/F/srchup

      $ pwd
      /tmp/A/B/C/D/E/F

      $ ./srchup thefile
      Found: /tmp/A/B/C/thefile


      We can see that the search went up the tree until it found what we were looking for.






      share|improve this answer

























        3












        3








        3







        A simple loop of checking the current directory and if it's not found then strip off the last component would work



        #!/bin/bash

        wantfile="$1"

        dir=$(realpath .)

        found=""

        while [ -z "$found" -a -n "$dir" ]
        do
        if [ -e "$dir/$wantfile" ]
        then
        found="$dir/$wantfile"
        fi
        dir=$dir%/*
        done

        if [ -z "$found" ]
        then
        echo Can not find: $wantfile
        else
        echo Found: $found
        fi


        For example, if this is the directory tree:



        $ find /tmp/A
        /tmp/A
        /tmp/A/good
        /tmp/A/good/show
        /tmp/A/good/show/this
        /tmp/A/B
        /tmp/A/B/C
        /tmp/A/B/C/thefile
        /tmp/A/B/C/D
        /tmp/A/B/C/D/E
        /tmp/A/B/C/D/E/F
        /tmp/A/B/C/D/E/F/srchup

        $ pwd
        /tmp/A/B/C/D/E/F

        $ ./srchup thefile
        Found: /tmp/A/B/C/thefile


        We can see that the search went up the tree until it found what we were looking for.






        share|improve this answer













        A simple loop of checking the current directory and if it's not found then strip off the last component would work



        #!/bin/bash

        wantfile="$1"

        dir=$(realpath .)

        found=""

        while [ -z "$found" -a -n "$dir" ]
        do
        if [ -e "$dir/$wantfile" ]
        then
        found="$dir/$wantfile"
        fi
        dir=$dir%/*
        done

        if [ -z "$found" ]
        then
        echo Can not find: $wantfile
        else
        echo Found: $found
        fi


        For example, if this is the directory tree:



        $ find /tmp/A
        /tmp/A
        /tmp/A/good
        /tmp/A/good/show
        /tmp/A/good/show/this
        /tmp/A/B
        /tmp/A/B/C
        /tmp/A/B/C/thefile
        /tmp/A/B/C/D
        /tmp/A/B/C/D/E
        /tmp/A/B/C/D/E/F
        /tmp/A/B/C/D/E/F/srchup

        $ pwd
        /tmp/A/B/C/D/E/F

        $ ./srchup thefile
        Found: /tmp/A/B/C/thefile


        We can see that the search went up the tree until it found what we were looking for.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jul 1 '16 at 23:20









        Stephen HarrisStephen Harris

        25.9k24477




        25.9k24477





















            3














            One way to do it:



            #! /bin/sh
            dir=$(pwd -P)
            while [ -n "$dir" -a ! -f "$dir/$1" ]; do
            dir=$dir%/*
            done
            if [ -f "$dir/$1" ]; then printf '%sn' "$dir/$1"; fi


            Replace pwd -P by pwd -L if you want to follow symlinks instead of checking physical directories.






            share|improve this answer























            • Any reason you use -a? [ -n "$dir" ] && ! [ -f "$dir/$1" ] is more reliable. (Exercise: Try determining correct parsing for [ -n "$a" -a "$b" = "$c" ] if a='=' and b='-o'.)

              – Wildcard
              Jul 2 '16 at 5:19











            • @Wildcard Because I'm an old fart and set in my bad old ways. :) Also, because it avoids running a second [ on shells where [ is not a builtin. As for [ -n "$a" -a "$b" = "$c" ], that's why you always write x"$b" = x"$c" rather than just "$b" = "$c".

              – Satō Katsura
              Jul 2 '16 at 5:30
















            3














            One way to do it:



            #! /bin/sh
            dir=$(pwd -P)
            while [ -n "$dir" -a ! -f "$dir/$1" ]; do
            dir=$dir%/*
            done
            if [ -f "$dir/$1" ]; then printf '%sn' "$dir/$1"; fi


            Replace pwd -P by pwd -L if you want to follow symlinks instead of checking physical directories.






            share|improve this answer























            • Any reason you use -a? [ -n "$dir" ] && ! [ -f "$dir/$1" ] is more reliable. (Exercise: Try determining correct parsing for [ -n "$a" -a "$b" = "$c" ] if a='=' and b='-o'.)

              – Wildcard
              Jul 2 '16 at 5:19











            • @Wildcard Because I'm an old fart and set in my bad old ways. :) Also, because it avoids running a second [ on shells where [ is not a builtin. As for [ -n "$a" -a "$b" = "$c" ], that's why you always write x"$b" = x"$c" rather than just "$b" = "$c".

              – Satō Katsura
              Jul 2 '16 at 5:30














            3












            3








            3







            One way to do it:



            #! /bin/sh
            dir=$(pwd -P)
            while [ -n "$dir" -a ! -f "$dir/$1" ]; do
            dir=$dir%/*
            done
            if [ -f "$dir/$1" ]; then printf '%sn' "$dir/$1"; fi


            Replace pwd -P by pwd -L if you want to follow symlinks instead of checking physical directories.






            share|improve this answer













            One way to do it:



            #! /bin/sh
            dir=$(pwd -P)
            while [ -n "$dir" -a ! -f "$dir/$1" ]; do
            dir=$dir%/*
            done
            if [ -f "$dir/$1" ]; then printf '%sn' "$dir/$1"; fi


            Replace pwd -P by pwd -L if you want to follow symlinks instead of checking physical directories.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Jul 2 '16 at 4:57









            Satō KatsuraSatō Katsura

            11k11634




            11k11634












            • Any reason you use -a? [ -n "$dir" ] && ! [ -f "$dir/$1" ] is more reliable. (Exercise: Try determining correct parsing for [ -n "$a" -a "$b" = "$c" ] if a='=' and b='-o'.)

              – Wildcard
              Jul 2 '16 at 5:19











            • @Wildcard Because I'm an old fart and set in my bad old ways. :) Also, because it avoids running a second [ on shells where [ is not a builtin. As for [ -n "$a" -a "$b" = "$c" ], that's why you always write x"$b" = x"$c" rather than just "$b" = "$c".

              – Satō Katsura
              Jul 2 '16 at 5:30


















            • Any reason you use -a? [ -n "$dir" ] && ! [ -f "$dir/$1" ] is more reliable. (Exercise: Try determining correct parsing for [ -n "$a" -a "$b" = "$c" ] if a='=' and b='-o'.)

              – Wildcard
              Jul 2 '16 at 5:19











            • @Wildcard Because I'm an old fart and set in my bad old ways. :) Also, because it avoids running a second [ on shells where [ is not a builtin. As for [ -n "$a" -a "$b" = "$c" ], that's why you always write x"$b" = x"$c" rather than just "$b" = "$c".

              – Satō Katsura
              Jul 2 '16 at 5:30

















            Any reason you use -a? [ -n "$dir" ] && ! [ -f "$dir/$1" ] is more reliable. (Exercise: Try determining correct parsing for [ -n "$a" -a "$b" = "$c" ] if a='=' and b='-o'.)

            – Wildcard
            Jul 2 '16 at 5:19





            Any reason you use -a? [ -n "$dir" ] && ! [ -f "$dir/$1" ] is more reliable. (Exercise: Try determining correct parsing for [ -n "$a" -a "$b" = "$c" ] if a='=' and b='-o'.)

            – Wildcard
            Jul 2 '16 at 5:19













            @Wildcard Because I'm an old fart and set in my bad old ways. :) Also, because it avoids running a second [ on shells where [ is not a builtin. As for [ -n "$a" -a "$b" = "$c" ], that's why you always write x"$b" = x"$c" rather than just "$b" = "$c".

            – Satō Katsura
            Jul 2 '16 at 5:30






            @Wildcard Because I'm an old fart and set in my bad old ways. :) Also, because it avoids running a second [ on shells where [ is not a builtin. As for [ -n "$a" -a "$b" = "$c" ], that's why you always write x"$b" = x"$c" rather than just "$b" = "$c".

            – Satō Katsura
            Jul 2 '16 at 5:30


















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